If we have two standard Gaussians with correlation $\rho,$ can we say something about the correlation of the events in which one gaussian is positive and the other is negative? Or both positive?
We would like a result that says that this correlation is bounded by $f(\rho),$ for some more or less explicit function $f$ that goes to $0$ when $\rho$ goes to $0.$
This correlation is $$f(\rho):=-\frac{2 }{\pi }\,\sin^{-1}\rho. \tag{0}\label{0}$$
Here is the graph $\{(\rho,f(\rho))\colon-1<\rho<1\}$:
One way to get \eqref{0} is as follows. Let $X,Y$ be standard jointly normal random variables with correlation $\rho$. Then the correlation in question is $$f(\rho)=\frac{P(X>0>Y)-P(X>0)P(Y<0)}{\sqrt{P(X>0)-P(X>0)^2} \sqrt{P(Y<0)-P(Y<0)^2}} \\ =4P(X>0>Y)-1, \tag{1}\label{1}$$ since $P(X>0)=1/2=P(Y<0)$. Next, without loss of generality $Y=\rho X+\sqrt{1-\rho^2}\,Z$, where $Z$ is a standard normal random variable independent of $X$. (This can be seen by computing the mean and the covariance matrix of $(X,\rho X+\sqrt{1-\rho^2}\,Z)$.) So, $$P(X>0>Y)=P(X>0,\rho X+\sqrt{1-\rho^2}\,Z<0) \\ =P\Big(X>0,Z<-\frac\rho{\sqrt{1-\rho^2}} X\Big),$$ which is the probability that the random point $(X,Z)$ on a plane will be between two rays emanating from the origin with angle $$\theta:=\cot^{-1}\frac\rho{\sqrt{1-\rho^2}} \tag{2}\label{2}$$ between them. Therefore and because the distribution of the random point $(X,Z)$ is rotation invariant, we have $$P(X>0>Y)=\frac\theta{2\pi}. \tag{3}\label{3}$$ Now \eqref{0} follows from \eqref{1}, \eqref{3}, and \eqref{2}. $\quad\Box$
